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Mathematics
Problem
13. Find the shortest distance between the lines
\nfrac
x
+
1
7
=
y
+
1
−
6
=
z
+
1
1
\nfrac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}
\nfrac
x
+
1
7
=
−
6
y
+
1
=
1
z
+
1
and
x
−
3
1
=
y
−
5
−
2
=
z
−
7
1
\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}
1
x
−
3
=
−
2
y
−
5
=
1
z
−
7
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